To solve the equation \( 2\sqrt{x} - 5 = 2 \), we will first isolate the square root term.
- Add 5 to both sides of the equation:
\[ 2\sqrt{x} = 2 + 5 \]
\[ 2\sqrt{x} = 7 \]
- Now divide both sides by 2:
\[ \sqrt{x} = \frac{7}{2} \]
- Next, square both sides to eliminate the square root:
\[ x = \left(\frac{7}{2}\right)^2 \]
\[ x = \frac{49}{4} \]
Now we have found a potential solution: \( x = \frac{49}{4} \).
Checking if this is an extraneous solution:
We need to substitute this value back into the original equation to verify that it satisfies the equation:
Original equation:
\[ 2\sqrt{x} - 5 = 2 \]
Substituting \( x = \frac{49}{4} \):
\[ 2\sqrt{\frac{49}{4}} - 5 = 2 \]
Calculate \( \sqrt{\frac{49}{4}} \):
\[ \sqrt{\frac{49}{4}} = \frac{7}{2} \]
Now substitute back into the equation:
\[ 2 \cdot \frac{7}{2} - 5 = 2 \]
This simplifies to:
\[ 7 - 5 = 2 \]
\[ 2 = 2 \]
Since both sides of the equation are equal, the solution \( x = \frac{49}{4} \) is valid and is not an extraneous solution.
Thus, the solution is:
\[ \boxed{\frac{49}{4}} \]